Prove that $\lfloor(n+1)a\rfloor-1$ is divisible by $(n+1)$ if $n= \left\lfloor \frac {1}{ a- \lfloor a \rfloor } \right\rfloor$

Question

I came across the following question across a math contest and was wondering how to solve it.

Let a be a positive real number that is not an integer and let

$$ n= \left\lfloor \frac {1}{ a- \lfloor a \rfloor } \right\rfloor $$

Prove that $\lfloor (n+1)a \rfloor -1 $ is divisible by $n+1$.

So I played around some values and got that that the quotient would be $\lfloor a \rfloor$. Would it be rigorous enough to prove that $\lfloor a \rfloor (n+1) = \lfloor (n+1)a \rfloor -1 $ if we have the above definition of $n$. Or would you recommend another approach?

Thanks.

Answer 1

The doubt seems to be due to the fact that the quotient $\lfloor a \rfloor $ comes from trying a few examples with literal numbers.

This procedure for coming up with $\lfloor a \rfloor $ is of course not a rigorous way of finding the quotient of the division by $n+1$, rather just a way to conjecture (really just to guess) what the quotient should be. But once you have such a guess, no matter how you guessed it, if you can then prove rigorously that $\lfloor a \rfloor (n+1) = \lfloor (n+1)a \rfloor -1 $ in all cases, not just in some cases, that is a perfectly rigorous proof that $\lfloor (n+1)a \rfloor - 1$ is divisible by $n + 1$.

Since your conjectured fact is true (I checked), I hope you will not have too much difficulty proving it.